In the given figure, △ A B C ∼ △ P Q R . Then, a r e a o f △ A B C a r e a o f △ P Q R equals Options: A . &nbsp A B 2 P Q 2 B . &nbsp B C 2 Q R 2 C . &nbsp A C 2 P R 2 D . &nbsp All of the above.

Answer: Option D : D We are given two triangles ABC and PQR such that △ A B C ∼ △ P Q R . For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below. Now, area of △ A B C = 1 2 × B C × A M and area of △ P Q R = 1 2 × Q R × P N So, a r e a o f △ A B C a r e a o f △ P Q R = 1 2 × B C × A M 1 2 × Q R × P N = B C × A M Q R × P N ⋯ ( 1 ) Now, in △ A B M and △ P Q N , ∠ B = ∠ Q (As △ A B C ∼ △ PQR) and ∠ A M B = ∠ P N Q = 90 ∘ . So, △ A B M ∼ △ P Q N ( By AA similarity criterion) ∴ A M P N = A B P Q ⋯ ( 2 ) Also, △ A B C ∼ △ P Q R (Given) So, A B P Q = A C P R = B C Q R ⋯ ( 3 ) From (1) and (3), we get, a r e a o f ( A B C ) a r e a o f ( P Q R ) = A B × A M P Q × P N = A B × A B P Q × P Q = A B 2 P Q 2 Now using (3), we get, a r e a o f △ A B C a r e a o f △ P Q R = A B 2 P Q 2 = B C 2 Q R 2 = A C 2 P R 2

In the given figure, △ABC∼△PQR. Then, area of △ ABCarea of △

Question

$In the given figure, △ A B C \sim △ P Q R . Then, \frac{a r e a o f △ A B C}{a r e a o f △ P Q R} equals$

Options:

A .

\frac{{A B}^{2}}{{P Q}^{2}}

B .

\frac{{B C}^{2}}{{Q R}^{2}}

C .

\frac{{A C}^{2}}{{P R}^{2}}

D . All of the above.

Answer: Option D
:
D

We are given two triangles ABC and PQR such that $△ A B C \sim △ P Q R$ .

For finding the areas of the two triangles, we draw altitudes AM and PN of the triangles ABC and PQR respectively, as shown below.

Now,
$area of △ A B C = \frac{1}{2} \times B C \times A M and area of △ P Q R = \frac{1}{2} \times Q R \times P N$
So,
$\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{\frac{1}{2} \times B C \times A M}{\frac{1}{2} \times Q R \times P N} = \frac{B C \times A M}{Q R \times P N} \dots (1)$
Now, in $△ A B M$ and $△ P Q N$ ,
$∠ B = ∠ Q$ (As $△ A B C \sim △$ PQR)
and $∠ A M B = ∠ P N Q = 90^{\circ}$ .
So, $△ A B M \sim △ P Q N$
( By AA similarity criterion)
$∴ \frac{A M}{P N} = \frac{A B}{P Q} \dots (2)$
Also, $△ A B C \sim △ P Q R$ (Given)
So, $\frac{A B}{P Q} = \frac{A C}{P R} = \frac{B C}{Q R} \dots (3)$
From (1) and (3), we get,
$\frac{a r e a o f (A B C)}{a r e a o f (P Q R)} = \frac{A B \times A M}{P Q \times P N} = \frac{A B \times A B}{P Q \times P Q} = \frac{A B^{2}}{P Q^{2}}$

Now using (3), we get,
$\frac{a r e a o f △ A B C}{a r e a o f △ P Q R} = \frac{A B^{2}}{P Q^{2}} = \frac{B C^{2}}{Q R^{2}} = \frac{A C^{2}}{P R^{2}}$